A $4$-choosable graph that is not $(8:2)$-choosable
Advances in Combinatronics (2019)
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In 1980, Erd\H{o}s, Rubin and Taylor asked whether for all positive integers
$a$, $b$, and $m$, every $(a:b)$-choosable graph is also $(am:bm)$-choosable.
We provide a negative answer by exhibiting a $4$-choosable graph that is not
$(8:2)$-choosable.
Publié le :
@article{ADVC_2019_a0,
author = {Zden\v{e}k Dvo\v{r}\'ak and Xiaolan Hu and Jean-S\'ebastien Sereni},
title = {A $4$-choosable graph that is not $(8:2)$-choosable},
journal = {Advances in Combinatronics},
publisher = {mathdoc},
year = {2019},
language = {en},
url = {http://geodesic.mathdoc.fr/item/ADVC_2019_a0/}
}
Zdeněk Dvořák; Xiaolan Hu; Jean-Sébastien Sereni. A $4$-choosable graph that is not $(8:2)$-choosable. Advances in Combinatronics (2019). http://geodesic.mathdoc.fr/item/ADVC_2019_a0/